Wize University Statics Textbook (Master) > Introduction

By looking at just the physical quantities, ignoring the numbers and the units, we are doing dimensional analysis. There are symbols that we can represent each physical quantity with, and we use square brackets to show that we are talking about the dimension. Here is a list of dimensions, notice how similar they are with units:

In dimensional analysis, we expect our equations and statements in physics to be dimensionally consistent. 
If I asked you: "What's the temperature like outside?" and you answered, "42 metres," I would say, "Wow! That's dimensionally inconsistent!"
In an equation, we would expect both sides to have the same dimensions in order for the equation to be dimensionally consistent. 

Example

Is this equation dimensionally consistent? (ccc is the speed of light)
E=mc2E = mc^2E=mc2
Answer is yes!

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Let's use dimensional analysis to answer the following problems:

1. You are in the last few minutes of an exam and you are rushed. You forget the exact formula for centripetal acceleration. You remember a few options of what it might look like. Which formula is correct?

a=vr         a=vr2        a=v2r      a=v2r2a=\frac{v}{r}\ \ \ \ \ \ \ \ \ a=\frac{v}{r^2}\ \ \ \ \ \ \ \ a=\frac{v^2}{r}\ \ \ \ \ \ a=\frac{v^2}{r^2}a=rv​         a=r2v​        a=rv2​      a=r2v2​

Write out the units of velocity, m/s, and r, s, for each option. Square them in the options where the quantities are squared. We need to have a term that gives us units of acceleration, ms2.\frac{m}{s^2}.s2m​.

Only the third option gives us this: a=v2r  →(m/s)2(m)=m2ms2=ms2a=\frac{v^2}{r}\ \ \rightarrow \frac{(m/s)^2}{(m)}=\frac{m^2}{m s^2}=\frac{m}{s^2}a=rv2​  →(m)(m/s)2​=ms2m2​=s2m​

Based on units, the correct formula is the third choice
.

2. You re-arrange a kinematic formula into the following form. Use dimensional analysis to determine if you successfully re-arranged the formula.

vf−vo2a=Δx\frac{v_f-v_o}{2a}=\Delta x2avf​−vo​​=Δx

Considering the units (velocity on the numerator, acceleration on the denominator):

m/sm/s2=s≠m\frac{m/s}{m/s^2}=s\neq mm/s2m/s​=s=m
vf−vo2a=Δx\frac{v_f-v_o}{2a}=\Delta x2avf​−vo​​=Δx
We wanted units of distance (m). So this is not the correct formula. (The correct formula is vf2−vo22a=Δx\frac{v_f^2-v_o^2}{2a}=\Delta x2avf2​−vo2​​=Δx).

Based on units, this IS NOT
 (is/is not) a correct formula.

3. Show that the following two terms use the same units: (PS, g has units of acceleration, m/s2)

12mv2          mgh\frac{1}{2}mv^2\ \ \ \ \ \ \ \ \ \ mgh21​mv2          mgh

Units of the left-hand term: (careful: "m" in a formula is a mass. "m" as a unit is a distance)

(kg)(m/s)2=kg m2s2(kg)(m/s)^2=\frac{kg~ m^2}{s^2}(kg)(m/s)2=s2kg m2​
Units of the right-hand term:

(kg)(m/s2)(m)=kg m2s2(kg)(m/s^2)(m)=\frac{kg~ m^2}{s^2}(kg)(m/s2)(m)=s2kg m2​
These are the same, so they have the same units. So we could add these terms together! By the way, this unit is called the Joule (J), and these terms describe energy.

X=YZ32WnX = \frac{Y Z}{32 W^n}X=32WnYZ​

If X=[Mθ]X = [M\theta]X=[Mθ] , Y=[MT2]Y = [MT^2]Y=[MT2] , Z=[θ/L2]Z = [\theta / L^2]Z=[θ/L2] , and W=[T/L]W = [T/L]W=[T/L] , what is the exponent n?